Consider a ladder sliding down a wall as in the figure. The variable a is the length of the ladder. The variable h is the height of the ladder's top at time t, and x is the distance from the wall to the ladder's bottom. Suppose that the length of the ladder is 4.6 meters and the top is sliding down the wall at a rate of 0.6m/s Calculate (dx)/(dt) when h=3.3 (Use decimal notation. Give your answer to three decimal places.) (dx)/(dt)vert _(vert xvert 5)approx square _(x,8)

To solve this problem, we will use the **Pythagorean theorem** and related rates. Here's the step-by-step solution:<br /><br />---<br /><br />### Step 1: Relating \( h \), \( x \), and \( a \)<br />The ladder forms a right triangle with the wall and the ground. The Pythagorean theorem gives us the relationship:<br />\[<br />h^2 + x^2 = a^2<br />\]<br />where:<br />- \( h \) is the height of the ladder's top from the ground,<br />- \( x \) is the distance of the ladder's bottom from the wall,<br />- \( a \) is the length of the ladder (constant).<br /><br />Since the ladder's length is constant (\( a = 4.6 \)), we have:<br />\[<br />h^2 + x^2 = 4.6^2<br />\]<br />\[<br />h^2 + x^2 = 21.16<br />\]<br /><br />---<br /><br />### Step 2: Differentiate with respect to time \( t \)<br />Differentiate both sides of the equation \( h^2 + x^2 = 21.16 \) with respect to \( t \):<br />\[<br />2h \frac{dh}{dt} + 2x \frac{dx}{dt} = 0<br />\]<br /><br />Simplify:<br />\[<br />h \frac{dh}{dt} + x \frac{dx}{dt} = 0<br />\]<br /><br />Solve for \( \frac{dx}{dt} \):<br />\[<br />\frac{dx}{dt} = -\frac{h}{x} \frac{dh}{dt}<br />\]<br /><br />---<br /><br />### Step 3: Substitute known values<br />We are given:<br />- \( a = 4.6 \),<br />- \( h = 3.3 \),<br />- \( \frac{dh}{dt} = -0.6 \) m/s (negative because the height is decreasing).<br /><br />First, calculate \( x \) when \( h = 3.3 \) using the Pythagorean theorem:<br />\[<br />h^2 + x^2 = 4.6^2<br />\]<br />\[<br />3.3^2 + x^2 = 21.16<br />\]<br />\[<br />10.89 + x^2 = 21.16<br />\]<br />\[<br />x^2 = 21.16 - 10.89<br />\]<br />\[<br />x^2 = 10.27<br />\]<br />\[<br />x = \sqrt{10.27} \approx 3.205 \, \text{m}<br />\]<br /><br />Now substitute \( h = 3.3 \), \( x \approx 3.205 \), and \( \frac{dh}{dt} = -0.6 \) into the formula for \( \frac{dx}{dt} \):<br />\[<br />\frac{dx}{dt} = -\frac{h}{x} \frac{dh}{dt}<br />\]<br />\[<br />\frac{dx}{dt} = -\frac{3.3}{3.205} (-0.6)<br />\]<br />\[<br />\frac{dx}{dt} \approx \frac{3.3}{3.205} \cdot 0.6<br />\]<br />\[<br />\frac{dx}{dt} \approx 0.617 \, \text{m/s}<br />\]<br /><br />---<br /><br />### Final Answer:<br />\[<br />\frac{dx}{dt} \approx 0.617 \, \text{m/s}<br />\]